What is the derivative of (x+6/(x^3+x-1)) ?

The derivative of (x+6/(x^3+x-1)) is (-2x^3-18x^2-7)/((x^3+x-1)^2)

What is the first derivative of (x+6/(x^3+x-1)) ?

The first derivative of (x+6/(x^3+x-1)) is (-2x^3-18x^2-7)/((x^3+x-1)^2)

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\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 1})

\frac{- 2 x ^{3} - 18 x ^{2} - 7}{( x ^{3} + x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x + 6 ) ( x ^{3} + x - 1 ) - \frac{d}{d x} ( x ^{3} + x - 1 ) ( x + 6 )}{( x ^{3} + x - 1 ) ^{2}}

\frac{d}{d x} (x + 6) = 1

\frac{d}{d x} (x^{3} + x - 1) = 3 x^{2} + 1

= \frac{1 \cdot ( x ^{3} + x - 1 ) - ( 3 x ^{2} + 1 ) ( x + 6 )}{( x ^{3} + x - 1 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{3} + x - 1 ) - ( 3 x ^{2} + 1 ) ( x + 6 )}{( x ^{3} + x - 1 ) ^{2}} : \frac{- 2 x ^{3} - 18 x ^{2} - 7}{( x ^{3} + x - 1 ) ^{2}}

= \frac{- 2 x ^{3} - 18 x ^{2} - 7}{( x ^{3} + x - 1 ) ^{2}}

y^{^{''}} + 4 y^{^{'}} + 3 y = t e^{- t}

x \to 6 lim (\frac{3 x}{x - 6})

\int r^{2} e^{r} d r

\frac{d}{d x} (cos (arctan (x)))

\int 6 e^{2 x} d x

What is the derivative of (x+6/(x^3+x-1)) ?
The derivative of (x+6/(x^3+x-1)) is (-2x^3-18x^2-7)/((x^3+x-1)^2)
What is the first derivative of (x+6/(x^3+x-1)) ?
The first derivative of (x+6/(x^3+x-1)) is (-2x^3-18x^2-7)/((x^3+x-1)^2)